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Thread: Critical numbers problem

  1. #1
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    Critical numbers problem

    Find all the critical numbers of the function f(x) = x+ln(|cos(x)|) that lie in the interval [pi, 3pi].

    Please help, thanks in advance.
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  2. #2
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    Re: Critical numbers problem

    Hello, nubshat!

    Exactly where is your difficulty?


    $\displaystyle \text{Find all the critical numbers of the function: }\:f(x) \:=\: x\;+\;\ln(|\cos x|)\:\text{ in the interval }[\pi, 3\pi].$

    Differentiate and equate to zero: .$\displaystyle f'(x) \:=\:1 + \frac{-\sin x}{\cos x} \:=\:0$

    We have: .$\displaystyle 1 - \tan x \:=\:0 \quad\Rightarrow\quad \tan x \:=\:1$

    Hence: .$\displaystyle x\:=\:\pm\tfrac{\pi}{4},\:\pm\tfrac{5\pi}{4},\:\pm \tfrac{9 \pi}{4},\:\pm\tfrac{13\pi}{4}\:\hdots$

    Answers: .$\displaystyle x \;=\;\frac{5\pi}{4},\:\frac{9\pi}{4}$
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  3. #3
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    Re: Critical numbers problem

    Doesn't the derivative fail to exist at $\displaystyle \frac{3\pi}{2}$ and $\displaystyle \frac{5\pi}{2}$, so that they are critical points, too?

    - Hollywood
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  4. #4
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    Re: Critical numbers problem

    would you also have to include the two end points at pi and 3pi?
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  5. #5
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    Re: Critical numbers problem

    Yes, I think both end points, too. Critical points are only used for finding minima and maxima, so assuming you're finding the minimum or maximum over the range $\displaystyle \pi$ to $\displaystyle 3\pi$, then you would need to include those points, too.

    I suppose it could be that you're finding the minimum or maximum over a different range, and for some reason you're temporarily limiting yourself to finding critical points in the given range. That doesn't make much sense, though.

    - Hollywood
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